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3 years ago

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You can increase the capacitance of a capacitor by A. Decreasing the plate spacing B. Increasing the plate spacing. ° C. Decreas

ing the area of the plates. D. Increasing the area of the plates. E. Both A and D F. Both B and C

1 answer:

eimsori [14]3 years ago

4 0

You can increase the capacitance of a capacitor by decreasing the plate spacing (A) or by increasing the area of the plates (D).

'A' and 'D' both do the job, so the correct choice is<em> (E)</em> .

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A square is 10 cm by 10 cm a student measure inside of the square is 9.9 cm and uses the measurements to calculate the area of t

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Answer:1.5 2.15

Explanation:

2,5

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A bowling ball with a mass of 7.0kg strikes a pin that had a mass of 2.0kg the pin flies forward with a velocity of 6.0m/s, and

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The conservation of momentum P states that the amount of momentum remains constant when there are not external forces.

We don't have external forces, so:

$P_0 = P_1\\m_bv_{0b}+m_pv_{0p}=m_bv_{1b}+m_pv_{1p}\\$

Where:

mb is the mass of the bowling ball
mp the mass of the pin
$v_{0b}\quad and\quad v_{0p}$ the initial velocities of the bowling ball and the pin.
$v_{1b}\quad and\quad v_{1p}$ the final velocities of the bowling ball and the pin.

Solving for v0b:

$v_{0b} =\dfrac{m_bv_{1b}+m_pv_{1p}- m_pv_{0p}}{m_{b}}\\\\v_{0b} =\dfrac{(7\;kg)(4\;m/s)+(2\;kg)(6\;m/s)- (2\;kg)(0 \;m/s)}{7\;kg}\\v_{0b}=\dfrac{40}{7}\;m/s\\\\\boxed{v_{0b}\approx5.71\;m/s}$

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3 years ago

A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to

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Answer:

1) $\Delta s=1000\ ft$

2) $\Delta s'=998.11\ ft.s^{-1}$

3) $t\approx125\ s$

$t'\approx463.733\ s$

Explanation:

Given:

width of river, $w=500\ ft$

speed of stream with respect to the ground, $v_s=8\ ft.s^{-1}$

speed of the swimmer with respect to water, $v=4\ ft.s^{-1}$

<u>Now the resultant of the two velocities perpendicular to each other:</u>

$v_r=\sqrt{v^2+v_s^2}$

$v_r=\sqrt{4^2+8^2}$

$v_r=8.9442\ ft.s^{-1}$

<u>Now the angle of the resultant velocity form the vertical:</u>

$\tan\beta=\frac{v_s}{v}$

$\tan\beta=\frac{8}{4}$

$\beta=63.43^{\circ}$

Now the distance swam by the swimmer in this direction be d.

so,

$d.\cos\beta=w$

$d\times \cos\ 63.43=500$

$d=1118.034\ ft$

Now the distance swept downward:

$\Delta s=\sqrt{d^2-w^2}$

$\Delta s=\sqrt{1118.034^2-500^2}$

$\Delta s=1000\ ft$

2)

On swimming 37° upstream:

<u>The velocity component of stream cancelled by the swimmer:</u>

$v'=v.\cos37$

$v'=4\times \cos37$

$v'=3.1945\ ft.s^{-1}$

<u>Now the net effective speed of stream sweeping the swimmer:</u>

$v_n=v_s-v'$

$v_n=8-3.1945$

$v_n=4.8055\ ft.s^{-1}$

<u>The component of swimmer's velocity heading directly towards the opposite bank:</u>

$v'_r=v.\sin37$

$v'_r=4\sin37$

$v'_r=2.4073\ ft.s^{-1}$

<u>Now the angle of the resultant velocity of the swimmer from the normal to the stream</u>:

$\tan\phi=\frac{v_n}{v'_r}$

$\tan\phi=\frac{4.8055}{2.4073}$

$\phi=63.39^{\circ}$

Now let the distance swam in this direction be d'.

$d'\times \cos\phi=w$

$d'=\frac{500}{\cos63.39}$

$d'=1116.344\ ft$

<u>Now the distance swept downstream:</u>

$\Delta s'=\sqrt{d'^2-w^2}$

$\Delta s'=\sqrt{1116.344^2-500^2}$

$\Delta s'=998.11\ ft.s^{-1}$

3)

Time taken in crossing the rive in case 1:

$t=\frac{d}{v_r}$

$t=\frac{1118.034}{8.9442}$

$t\approx125\ s$

Time taken in crossing the rive in case 2:

$t'=\frac{d'}{v'_r}$

$t'=\frac{1116.344}{2.4073}$

$t'\approx463.733\ s$

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